Answer: (i) \(\alpha^2+\beta^2+\gamma^2=2p\)
(ii) \(\alpha^3+\beta^3+\gamma^3=3q\)
(iii) \(6(\alpha^5+\beta^5+\gamma^5)=5(\alpha^3+\beta^3+\gamma^3)(\alpha^2+\beta^2+\gamma^2)\)
Since \(x^3-px-q=0\) has roots \(\alpha,\beta,\gamma\), we have, by comparing coefficients,
\(\alpha+\beta+\gamma=0,\\qquad \alpha\beta+\beta\gamma+\gamma\alpha=-p,\\qquad \alpha\beta\gamma=q.\)
(i) Use the identity
\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha).\)
Substituting the values above gives
\(\alpha^2+\beta^2+\gamma^2=0-2(-p)=2p.\)
(ii) For three numbers,
\(\alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)+3\alpha\beta\gamma.\)
As \(\alpha+\beta+\gamma=0\), this simplifies to
\(\alpha^3+\beta^3+\gamma^3=3\alpha\beta\gamma=3q.\)
(iii) For the fifth powers, use the standard recurrence for power sums coming from the cubic relation. Since each root satisfies \(r^3=pr+q\), multiplying by \(r^2\) gives \(r^5=pr^3+qr^2\). Summing over \(r=\alpha,\beta,\gamma\) gives
\(\alpha^5+\beta^5+\gamma^5=p(\alpha^3+\beta^3+\gamma^3)+q(\alpha^2+\beta^2+\gamma^2).\)
Now use parts (i) and (ii):
\(\alpha^5+\beta^5+\gamma^5=p(3q)+q(2p)=5pq.\)
Therefore
\(6(\alpha^5+\beta^5+\gamma^5)=6\cdot 5pq=30pq,\)
while
\(5(\alpha^3+\beta^3+\gamma^3)(\alpha^2+\beta^2+\gamma^2)=5(3q)(2p)=30pq.\)
Hence
\(6(\alpha^5+\beta^5+\gamma^5)=5(\alpha^3+\beta^3+\gamma^3)(\alpha^2+\beta^2+\gamma^2).\)
